Question

Find an equation of the tangent line to the curve at the given point (Calculus Derivatives).

Answer 1

Question #1

Answer 2

ok what did you find for y'?

Answer 3

\[\lim_{h \rightarrow 0}\frac{ f(a+h)-f(a) }{ h }\] = \[\lim_{h \rightarrow 0}\frac{ [(3+h)^3-3(3+h)+1-(3^3-3(3)+1)] }{ h }\] . Putting more work on here second

Answer 4

do you have to apply differentiation by first principles? :/

Answer 5

\[(3+h)^3; (3+h)(3+h)=9+3h+3h+h^2=9+6h+h^2 ; (9+6h+h^2)(3+h) = 27+18h+3h^2+9h+6h^2+h^3\]

Answer 6

Are you familiar with the power rule of differentiation?

\[\frac{ d }{ dx }[x^n]=n*x^{n-1}\]

Answer 7

you might like pascal's triangle 4th row says 1 3 3 1
\[(3+h)^3=1(3)^3+3(3)^2h+3(3)h^2+h^3\]

Answer 8

Our class isn't on there yet, this is the beginning of calculus for the end of Chapter 2

Answer 9

\[\lim_{h \rightarrow 0} \frac{3^3+3(3)^2h+3(3)h^2+h^3-3(3+h)+1-(3^3-3(3)+1)}{h}\]

Answer 10

Sec. I might found the answer.

Answer 11

I gotten 6x+1 as the answer.

Answer 12

ok let's go slow then you do see above the 3^3 's cancel right?

Answer 13

since 3^3-3^3=0

and also you should see the 3(3) also known as 9's cancel you know since -9+9=0

Answer 14

This is how the book does it. Similar problem.

Answer 15

you should also see the 1's cancel since 1-1=0

Answer 16

\[\lim_{h \rightarrow 0} \frac{3^3+3(3)^2h+3(3)h^2+h^3-3(3+h)+1-(3^3-3(3)+1)}{h} \\ \lim_{h \rightarrow 0} \frac{3(3)^2h+3(3)h^2+h^3-3h}{h}\]

Answer 17

now divide top and bottom by h

Answer 18

\[\lim_{h \rightarrow 0}(h^2+9h+27)\] Did u get that?

Answer 19

should wind up with
\[\lim_{h \rightarrow 0} (3(3)^2+3(3)h+h^2-3)\]

Answer 20

something happened to your -3 part above

Answer 21

should be
\[\lim_{h \rightarrow 0} (h^2+9h+27-3)\]

Answer 22

so your slope on your homework is off
which will also make your y-intercept you found off too

Answer 23

please let me know if I need to redo the algebra for you if you don't understand

Answer 24

I guess you can call it algebra/arithmetic whatever that stuff is

Answer 25

\[\lim_{h \rightarrow 0}(h^2+9h-24)\] that correct? I'm looking at calculations at the moment.

Answer 26

well 27-3 is 24 not -24 :p

Answer 27

By using the point (3,19) and using the point-slope form for slope M=24. I get \[y=24x+19\]

Answer 28

y-19=f'(3)(x-3)
y-19=24(x-3)
y-19=24x-72

hmmm...-72+19...

Answer 29

\[y=24x+53\]

Answer 30

I hate to be that guy but -72+19 is -53

Answer 31

Yea so the answer is \[y=24x-53\] ?

Answer 32

yep

Answer 33

do you understand how we got the slope to be 24?

Answer 34

Yea, turns out I was messing up on simple arithmetic/algebra breaks for: adding/subtracting/factoring etc.

Answer 35

Yea I understood the core concepts, just messed up with the things from above.

Answer 36

Time for a study break haha

Answer 37

Thanks a lot for sticking around and helping. :)

Answer 38

we all mess on arithmetic
just go pet a cat and then come back
you will be already
(that is my only study advice; cats for the win!)